EEVblog Electronics Community Forum
Electronics => Projects, Designs, and Technical Stuff => Topic started by: theleakydiode on July 16, 2015, 07:44:19 pm
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Hi, I am looking for a low value current sense resistor but am having trouble understanding the non standard resistor voltage ratings in the datasheet. For example the LVR03R0700FE12 is a 70 Milliohm part and the datasheet states that its maximum working voltage is (P x R) 1/2.
Would that make 105mV the max voltage rating or have I got that wrong? I intend to use two of them in parallel for 35 Milliohms and they would share a current of about 10 amps + peaks.
But I don't really understand if I am interpreting it correctly as such a low voltage seems off, any help would be appreciated.
Thanks.
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I get 458mV.
LVR03 so P = 3W
R = 0.07
V = (3*0.07)0.5 = 0.458V
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I get 458mV.
LVR03 so P = 3W
R = 0.07
V = (3*0.07)0.5 = 0.458V
How did you get that Hero999? 3x.07=0.21v which divided by 2 makes 105mV.
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I get 458mV.
LVR03 so P = 3W
R = 0.07
V = (3*0.07)0.5 = 0.458V
How did you get that Hero999? 3x.07=0.21v which divided by 2 makes 105mV.
That is not divided by 2 it is to the exponent of 1/2 (aka square root).
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I get 458mV.
LVR03 so P = 3W
R = 0.07
V = (3*0.07)0.5 = 0.458V
How did you get that Hero999? 3x.07=0.21v which divided by 2 makes 105mV.
That is not divided by 2 it is to the exponent of 1/2 (aka square root).
That explains it then. It seems like a low value, does this make them a bad choice for peak current sensing in a flyback converter? As the inductor peak current will be much higher than the average current.
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Twinkle, twinkle little star, power equals I squared - R.
From that we obtain, since V = I x R,
Power = V-squared over R.
(Try not to get confused with "little star up in the sky, power equals R squared I".)
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It seems like a low value, does this make them a bad choice for peak current sensing in a flyback converter? As the inductor peak current will be much higher than the average current.
It depends on on the rise/fall time of the current. At high frequencies using low resistor values the voltage generated due to the parasitic inductance distorts the actual measurement. I can't find any inductance value in the datasheet except the meaningless "Low inductance" features at the first page, but you can assume at least 10nH/cm length.
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It's interesting they've stating the maximum voltage rating, just by the amount of power the resistor can safely dissipate.
Most resistors can withstand momentary voltage surges which will exceed the power rating, so long as the limiting element voltage isn't exceeded. For example a metal film 1W 100R resistor may only be able to take 10V continuously but it can take brief 100V pulses, as long as the average power dissipated doesn't exceed 1W. However a 1kV pulse would cause it to arc over and be damaged.
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i think they are saying the maximum voltage u read off the resistor as a shunt element.
so v=i*r
r = 0.08 ohm
0.08*I = 0.418V
max voltage drop, not max voltage. thats an absurdely low max voltage
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max voltage drop, not max voltage. thats an absurdely low max voltage
A resistor only has two terminals, only the voltage drop across it could possibly be relevant? What are these two different limits of which you speak?
It's interesting they've stating the maximum voltage rating, just by the amount of power the resistor can safely dissipate.
Most resistors can withstand momentary voltage surges which will exceed the power rating, so long as the limiting element voltage isn't exceeded. For example a metal film 1W 100R resistor may only be able to take 10V continuously but it can take brief 100V pulses, as long as the average power dissipated doesn't exceed 1W. However a 1kV pulse would cause it to arc over and be damaged.
I dare say that in a resistor where the maximum continuous working voltage is less than a volt, any peaks from a flyback converter are going to be less than 10V, and evidently incapable of producing arcover. So I think the part will fairly obviously not get destroyed in the OPs case. However, I agree that it's poor that this isn't actually specified properly in the datasheet.
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OK, you've got a 70 milliohm, 3 watt device.
By my maths that means about 6.5A is the allowable current, or 13A since you're putting two in parallel.